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+# Math 4111 Final Review
+
+## Weierstrass M-test
+
+Let $\sum_{n=1}^{\infty} f_n(x)$ be a series of functions.
+
+The weierstrass M-test goes as follows:
+
+1. $\exists M_n \geq 0$ such that $\forall x\in E, |f_n(x)| \leq M_n$.
+2. $\sum M_n$ converges.
+
+Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly.
+
+Example:
+
+### Ver.0
+
+$\forall x\in [-1,1)$,
+
+$$
+\sum_{n=1}^{\infty} \frac{x^n}{n}
+$$
+
+converges. (point-wise convergence on $[-1,1)$)
+
+$\forall x\in [-1,1)$,
+
+$$
+\left| \frac{x^n}{n} \right| \leq \frac{1}{n}
+$$
+
+Since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we don't know if the series converges uniformly or not using the weierstrass M-test.
+
+### Ver.1
+
+However, if we consider the series on $[-1,1]$,
+
+$$
+\sum_{n=1}^{\infty} \frac{x^n}{n^2}
+$$
+
+converges uniformly. Let $M_n = \frac{1}{n^2}$. This satisfies the weierstrass M-test. And this series converges uniformly on $[-1,1]$.
+
+### Ver.2
+
+$\forall x\in [-\frac{1}{2},\frac{1}{2}]$,
+
+$$
+\sum_{n=1}^{\infty} \frac{x^n}{n}
+$$
+
+converges uniformly. Since $\left| \frac{x^n}{n} \right|=\frac{|x|^n}{n}\leq \frac{(1/2)^n}{n}\leq \frac{1}{2^n}=M_n$, by geometric series test, $\sum_{n=1}^{\infty} M_n$ converges.
+
+M-test still not applicable here.
+
+$$
+\sum_{n=1}^{\infty} \frac{x^n}{n}
+$$
+
+converges uniformly on $[-\frac{1}{2},\frac{1}{2}]$.
+
+> Comparison test:  
+> 
+> For a series $\sum_{n=1}^{\infty} a_n$, if
+> 
+> 1. $\exists M_n$ such that $|a_n|\leq M_n$
+> 2. $\sum_{n=1}^{\infty} M_n$ converges
+> 
+> Then $\sum_{n=1}^{\infty} a_n$ converges.
+
+## Proving continuity of a function
+
+If $f:E\to Y$ is continuous at $p\in E$, then for any $\epsilon>0$, there exists $\delta>0$ such that for any $x\in E$, if $|x-p|<\delta$, then $|f(x)-f(p)|<\epsilon$.
+
+Example:
+
+Let $f(x)=2x+1$. For $p=1$, prove that $f$ is continuous at $p$.
+
+Let $\epsilon>0$ be given. Let $\delta=\frac{\epsilon}{2}$. Then for any $x\in \mathbb{R}$, if $|x-1|<\delta$, then
+
+$$
+|f(x)-f(1)|=|2x+1-3|=|2x-2|=2|x-1|<2\delta=\epsilon.
+$$
+
+Therefore, $f$ is continuous at $p=1$.
+
+_You can also use smaller $\delta$ and we don't need to find the "optimal" $\delta$._
+
+## Play of open covers
+
+Example of non compact set:
+
+$\mathbb{Q}$ is not compact, we can construct an open cover $G_n=(-\infty,\sqrt{2})\cup (\sqrt{2}+\frac{1}{n},\infty)$.
+
+Every unbounded set is not compact, we construct an open cover $G_n=(-n,n)$.
+
+Every k-cell is compact.
+
+Every finite set is compact.
+
+Let $p\in A$ and $A$ is compact. Then $A\backslash \{p\}$ is not compact, we can construct an open cover $G_n=(\inf(A)-1,p)\cup (p+\frac{1}{n},\sup(A)+1)$.
+
+If $K$ is closed in $X$ and $X$ is compact, then $K$ is compact.
+
+Proof: 
+
+Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $K$.
+
+> $A$ is open in $X$, if and only if $X\backslash A$ is closed in $X$.
+
+Since $X\backslash K$ is opened in $X$, $\{G_\alpha\}_{\alpha\in A}\cup \{X\backslash K\}$ is an open cover of $X$.
+
+Since $X$ is compact, there exists a finite subcover $\{G_{\alpha_1},\cdots,G_{\alpha_n},X\backslash K\}$ of $X$.
+
+Since $X\backslash K$ is not in the subcover, $\{G_{\alpha_1},\cdots,G_{\alpha_n}\}$ is a finite subcover of $K$.
+
+Therefore, $K$ is compact.
+
+## Cauchy criterion
+
+### In sequences
+
+Def: A sequence $\{a_n\}$ is Cauchy if for any $\epsilon>0$, there exists $N$ such that for any $m,n\geq N$, $|a_m-a_n|<\epsilon$.
+
+Theorem: In $\mathbb{R}$, every sequence is Cauchy if and only if it is convergent.
+
+### In series
+
+Let $s_n=\sum_{k=1}^{n} a_k$.
+
+Def: A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $\{s_n\}$ converges.
+
+$\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
+
+$$
+|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
+$$
+
+## Comparison test
+
+If $|a_n|\leq b_n$ and $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.
+
+Proof:
+
+Since $\sum_{n=1}^{\infty} b_n$ converges, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
+
+$$
+\left|\sum_{k=n+1}^{m} b_k\right|<\epsilon.
+$$
+
+By triangle inequality,
+
+$$
+\left|\sum_{k=n}^{m}a_k\right|\leq \sum_{k=n+1}^{m} |a_k|\leq \sum_{k=n+1}^{m} b_k<\epsilon.
+$$
+
+Therefore, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
+
+$$
+|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
+$$
+
+Therefore, $\{s_n\}$ is Cauchy, and $\sum_{n=1}^{\infty} a_n$ converges.
+